Simplify the following expression: $y = \dfrac{-3x^2- 10x- 3}{-3x - 1}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-3)}{(-3)} &=& 9 \\ {a} + {b} &=& &=& {-10} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $9$ and add them together. The factors that add up to ${-10}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-1}$ and ${b}$ is ${-9}$ $ \begin{eqnarray} {ab} &=& ({-1})({-9}) &=& 9 \\ {a} + {b} &=& {-1} + {-9} &=& -10 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-3}x^2 {-1}x) + ({-9}x {-3}) $ Factor out the common factors: $ x(-3x - 1) + 3(-3x - 1)$ Now factor out $(-3x - 1)$ $ (-3x - 1)(x + 3)$ The original expression can therefore be written: $ \dfrac{(-3x - 1)(x + 3)}{-3x - 1}$ We are dividing by $-3x - 1$ , so $-3x - 1 \neq 0$ Therefore, $x \neq -\frac{1}{3}$ This leaves us with $x + 3; x \neq -\frac{1}{3}$.